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STRUCTURE OF SEABORGIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( October 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favor of various contradicting nuclear theories which cannot lead to the nuclear structure. For example for the interpretation of nuclear fission in the “Nuclear fission-WIKIPEDIA ” one reads the fallacious nuclear models along with the invalid relativity including the wrong mass-energy equivalence. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the electromagnetic laws. (See my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS ). Seaborgium (Sg) is an artificial element, and thus a standard atomic mass cannot be given. Like all artificial elements, it has no stable isotopes. The first isotope to be synthesized was Sg-263m in 1974. There are 12 known radioisotopes from Sg-258 to Sg-271 and 2 known isomers (Sg -261m and Sg-263m). The longest-lived isotope is 271Sg with a half-life of 2.4 minutes. It has 59 extra neutrons which fill all the 59 blank positions, but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. NUCLEAR STRUCTURE OF Sg ISOTOPES HAVING EVEN NUMBER OF EXTRA NEUTRONS It is well well-known that the structure of lead-164 (core) of high symmetry consists of 8 horizontal planes and 2 horizontal lines providing 44 blank positions for receiving extra neutrons with two bonds per neutron in order to construct the stable Pb-208. (See the fourth figure of lead at the bottom of the page). Similarly the structure of Sg-212 (core) with 106 protons and 106 neutrons (even number) consists of 8 horizontal planes of opposite spins, including four additional deuterons with S = +2 and S = -2 which exist over and under the structure of 8 horizontal planes, forming the up horizontal line (+UHL) and the down horizontal line (-DHL). So all these nucleons of the 8 horizontal planes and the +UHL and the -DHL give S = 0 . In general, the structure of Sg-212 (core) has S =0 and is similar to the structure of Pb-164, because the two additional vertical systems of p105n105 and p106n106 with S = 0 make symmetrical vertical rectangles. So all the nuclides with even number of extra neutrons existing from Sg-258 to Sg-266 are based on the Sg-212 with S =0, because their total spin S =0 is due to the extra neutrons of opposite spins. For example the Sg-258 has 46 extra neutrons with opposite spins. they fill 46 blank positions but the pp repulsions of long range (large number) always overcome such pn bonds leading to the decay. ' ' NUCLEAR STRUCTURE OF Sg-259 WITH S = +1/2 Also the structure of Sg-259 with S = +1/2 is based on the structure of Sg-212 (core) with S = 0. In this case the Sg-259 of 47 extra neutrons has 24 extra neutrons of positive spins and 23 extra neutrons of negative spins. That is S = 0 + 24(+1/2) + 23(-1/2) = +1/2 Here the 47 extra neutrons fill 47 blank positions, but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. ' ' NUCLEAR STRUCTURE OF Sg-261 WITH S = +7/2 After a careful analysis I found that the structure of such an unstable nuclide, with odd number of extra neutrons, is based on another structure of the Sg-212 (core) having S = +4. In this case the two deuterons of the -DHL change their spins from S = -2 to S = +2 giving S = +4, because they move to +UHL for making horizontal bonds with a deuteron of the up horizontal line. Under this condition the Sg-261 with S = +7/2 of 49 extra neutrons, has 24 extra neutrons of positive spins and 25 extra neutrons of negative spins . That is S = +4 + 24(+1/2) + 25(-1/2) = +7/2 Here the 49 extra neutrons fill 49 blank positions, but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. ' ' NUCLEAR STRUCTURE OF Sg-263 WITH S = +9/2 After a careful analysis I found that the structure of Sg-263 of 51 extra neutrons is based on another structure of Sg-212 having S = +4. In this case the two deuterons of the -DHL change their spins from S =-2 to S =+2 giving S = +4, because they move to +UHL for making horizontal bonds with the two deuterons of the up horizontal line. Note that Sg-263 of 51 extra neutrons has 26 extra neutrons of positive spins and 25 extra neutrons of negative spins. That is S = +4 + 23(+1/2) + 24(-1/2) = +9/2 Here the 51 extra neutrons fill 51 blank positions but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. Category:Fundamental physics concepts